3.655 \(\int \frac {-e^{-x}+e^x}{e^{-x}+e^x} \, dx\)

Optimal. Leaf size=10 \[ \log \left (e^{-x}+e^x\right ) \]

[Out]

ln(exp(-x)+exp(x))

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Rubi [A]  time = 0.04, antiderivative size = 12, normalized size of antiderivative = 1.20, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2282, 446, 72} \[ \log \left (e^{2 x}+1\right )-x \]

Antiderivative was successfully verified.

[In]

Int[(-E^(-x) + E^x)/(E^(-x) + E^x),x]

[Out]

-x + Log[1 + E^(2*x)]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {align*} \int \frac {-e^{-x}+e^x}{e^{-x}+e^x} \, dx &=\operatorname {Subst}\left (\int \frac {-1+x^2}{x \left (1+x^2\right )} \, dx,x,e^x\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {-1+x}{x (1+x)} \, dx,x,e^{2 x}\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {1}{x}+\frac {2}{1+x}\right ) \, dx,x,e^{2 x}\right )\\ &=-x+\log \left (1+e^{2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 12, normalized size = 1.20 \[ \log \left (e^{2 x}+1\right )-x \]

Antiderivative was successfully verified.

[In]

Integrate[(-E^(-x) + E^x)/(E^(-x) + E^x),x]

[Out]

-x + Log[1 + E^(2*x)]

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fricas [A]  time = 0.41, size = 11, normalized size = 1.10 \[ -x + \log \left (e^{\left (2 \, x\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1/exp(x)+exp(x))/(exp(-x)+exp(x)),x, algorithm="fricas")

[Out]

-x + log(e^(2*x) + 1)

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giac [A]  time = 0.21, size = 11, normalized size = 1.10 \[ -x + \log \left (e^{\left (2 \, x\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1/exp(x)+exp(x))/(exp(-x)+exp(x)),x, algorithm="giac")

[Out]

-x + log(e^(2*x) + 1)

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maple [A]  time = 0.04, size = 14, normalized size = 1.40 \[ \ln \left ({\mathrm e}^{2 x}+1\right )-\ln \left ({\mathrm e}^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1/exp(x)+exp(x))/(exp(-x)+exp(x)),x)

[Out]

ln(1+exp(x)^2)-ln(exp(x))

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maxima [A]  time = 0.88, size = 8, normalized size = 0.80 \[ \log \left (e^{\left (-x\right )} + e^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1/exp(x)+exp(x))/(exp(-x)+exp(x)),x, algorithm="maxima")

[Out]

log(e^(-x) + e^x)

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mupad [B]  time = 3.54, size = 11, normalized size = 1.10 \[ \ln \left ({\mathrm {e}}^{2\,x}+1\right )-x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x) - exp(x))/(exp(-x) + exp(x)),x)

[Out]

log(exp(2*x) + 1) - x

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sympy [A]  time = 0.10, size = 8, normalized size = 0.80 \[ - x + \log {\left (e^{2 x} + 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1/exp(x)+exp(x))/(exp(-x)+exp(x)),x)

[Out]

-x + log(exp(2*x) + 1)

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